Knapsack Secretary with Bursty Adversary

June 20, 2020 Β· Declared Dead Β· πŸ› International Colloquium on Automata, Languages and Programming

πŸ‘» CAUSE OF DEATH: Ghosted
No code link whatsoever

"No code URL or promise found in abstract"

Evidence collected by the PWNC Scanner

Authors Thomas Kesselheim, Marco Molinaro arXiv ID 2006.11607 Category cs.DS: Data Structures & Algorithms Citations 14 Venue International Colloquium on Automata, Languages and Programming Last Checked 3 months ago
Abstract
The random-order or secretary model is one of the most popular beyond-worst case model for online algorithms. While it avoids the pessimism of the traditional adversarial model, in practice we cannot expect the input to be presented in perfectly random order. This has motivated research on ``best of both worlds'' (algorithms with good performance on both purely stochastic and purely adversarial inputs), or even better, on inputs that are a mix of both stochastic and adversarial parts. Unfortunately the latter seems much harder to achieve and very few results of this type are known. Towards advancing our understanding of designing such robust algorithms, we propose a random-order model with bursts of adversarial time steps. The assumption of burstiness of unexpected patterns is reasonable in many contexts, since changes (e.g. spike in a demand for a good) are often triggered by a common external event. We then consider the Knapsack Secretary problem in this model: there is a knapsack of size $k$ (e.g., available quantity of a good), and in each of the $n$ time steps an item comes with its value and size in $[0,1]$ and the algorithm needs to make an irrevocable decision whether to accept or reject the item. We design an algorithm that gives an approximation of $1 - \tilde{O}(Ξ“/k)$ when the adversarial time steps can be covered by $Ξ“\ge \sqrt{k}$ intervals of size $\tilde{O}(\frac{n}{k})$. In particular, setting $Ξ“= \sqrt{k}$ gives a $(1 - O(\frac{\ln^2 k}{\sqrt{k}}))$-approximation that is resistant to up to a $\frac{\ln^2 k}{\sqrt{k}}$-fraction of the items being adversarial, which is almost optimal even in the absence of adversarial items. Also, setting $Ξ“= \tildeΞ©(k)$ gives a constant approximation that is resistant to up to a constant fraction of items being adversarial.
Community shame:
Not yet rated
Community Contributions

Found the code? Know the venue? Think something is wrong? Let us know!

πŸ“œ Similar Papers

In the same crypt β€” Data Structures & Algorithms

Died the same way β€” πŸ‘» Ghosted